The Riemann-Liouville fractional integral of order $\alpha >0$ for a continuous function f is defined as provided that the right-hand-side integral exists, where $\mathsf{\Gamma }\left(\alpha \right)$ denotes the Gamma function is the Euler gamma function defined by

Let $n\in \mathbb{N}$ be a positive integer and α be a positive real such that $n-1<\alpha \le n$, then the fractional derivative of a function f in the Caputo sense is defined as provided that the right-hand-side integral exists and is finite. We notice that the Caputo derivative of a constant is zero.

Let α and β be positive reals. If f is a continuous function, then we have

Let α be positive real. Then we have

Let $n\in \mathbb{N}$ and $n-1<\alpha \le n$. If u is a continuous function, then we have

If $h\in C[0,1]$, then the unique solution of the boundary value problem (1.3) and (1.2) is given by where

From Lemmas 2.1, 2.2 and 2.3 and the Definition 2.1, it follows that and By inserting the boundary condition $u\left(0\right)=0$ in (2.2) gives $c_2=0$ and also by inserting the boundary condition ${}^{c}Du\left(0\right)=0$ in (2.1) gives $c_0=0$. Inserting (2.2) into (2.1) to obtain Using the third boundary condition in (1.2) gives Substituting the above values in (2.2) to obtain the desired results. □

If $\lambda =\mathsf{\Gamma }(\alpha +2)$ and ${\beta }_i$ are positive (negative) for all $i\in \mathbb{N}$, then we have

([22,23]). Let $\mathbb{E}$ be a Banach space, C be a closed and convex subset of $\mathbb{E}$, U be an open subset of C and $0\in U$. Suppose that the operator $\mathcal{T}:\overline{U}\to C$ is a continuous and compact map (that is, $\mathcal{T}\left(\overline{U}\right)$ is a relatively compact subset of C). Then either

Let ${\mathcal{B}}_r=\{u\in \mathbb{E}:\parallel u\parallel \le r\}$ be a closed ball with the radius $r\ge \mathcal{M}\mathbf{A}/(1-\mathbf{Q})$ where Then, for $u\in {\mathcal{B}}_r$, we have From this, we obtain Whence, we have which leads to $Tu\subset {\mathcal{B}}_r$. Now, let $u,v\in {\mathcal{B}}_r$, then we have Therefore, we have By the hypothesis $\mathbf{Q}<1$, it follows that the operator T defined in (3.3) is a contraction. Therefore, with Banach contraction mapping principle, we deduce that the operator T has a fixed point, which equivalently implies that the boundary value problem (1.1)-(1.2) has a unique solution on $[0,1]$. □

Assume that the assumptions (${\mathcal{H}}_1$) and (${\mathcal{H}}_3$) hold. Then the boundary value problem (1)-(2) has at least one solution if there exists a constant $M>0$ such that $\mathbf{K}>1$ where $\mathbf{K}$ is given

The continuity of the function f implies that the operator $T:\mathbb{E}\to \mathbb{E}$ defined by (3.3) is continuous. Assume that ${\mathcal{B}}_r=\{u\in \mathbb{E}:\parallel u\parallel <r\}$ be an open subset of the Banach space $\mathbb{E}$ with radius $r>0$. First, we are in a position to prove that the operator $T:\mathbb{E}\to \mathbb{E}$ is completely continuous. Assume that $u\in {\mathcal{B}}_r$. Then, we have It follows that which concludes the boundedness of the operator T. Suppose that $t_1,t_2\in [0,1]$ such that $t_1<t_2$, it follows that It is clear that the right-hand side of the above inequality approaches zero independently of $u\in \mathbb{E}$ as $t_1\to t_2$. Since the operator T satisfies the above assumptions, it follows by the Arzela-Ascoli theorem that $T:\mathbb{E}\to \mathbb{E}$ is completely continuous.

According to the Leray-Schauder nonlinear alternative Lemma 2.5, the result will follow once we prove the boundedness of the set of all solutions to equations $u=\delta Tu$ for some $\delta \in [0,1]$. Let u is a solution of the equation $u=\delta Tu$ for some $\delta \in [0,1]$, then for all $t\in [0,1]$, from the boundedness of the operator T, we have which implies that By the assumption $\mathbf{K}>1$, then there exists a constant $M>0$ such that $\parallel u\parallel \ne M$. Setting the open set Based on the form of $\mathsf{\Omega }$, there is no $u\in \partial \mathsf{\Omega }$ such that $u=\delta T\left(u\right)$ for some $\delta \in (0,1)$. Since the operator $T:\overline{\mathsf{\Omega }}\to \mathbb{E}$ is continuous and completely continuous, then by the nonlinear alternative of Leray-Schauder type Lemma 2.5, we deduce that T has a fixed point $u\in \overline{\mathsf{\Omega }}$ which is a solution of problem (1.1)-(1.2). This ends the proof. □

Assume that the assumptions (${\mathcal{H}}_1$) and (${\mathcal{H}}_4$) hold. Then the boundary value problem (1)-(2) has at least one solution if where $\mathbf{A}$ and $\mathbf{B}$ are given by (3.1) and (3.2), respectively.

Let us define the open ball ${\mathcal{B}}_r\subset \mathbb{E}$ with radius $r>0$ as where r will be determined later. It is adequate to prove that the operator $T:\overline{{\mathcal{B}}_r}\to \mathbb{E}$ satisfies To do this, assume that $u=\sigma Tu$ for some $\sigma \in [0,1]$. Then, as in the preceding results, we have which implies that provided that $\eta \mathbf{A}+\mathbf{B}<1$ which leads to $\eta <(1-\mathbf{B})/\mathbf{A}$. Now, suppose that there exists $ϵ>0$ such that By the analysis above, it follows that the relation (3.4) holds. Let us now define the continuous operator In view of the results in Theorems above, it is clear that the operator $h_{\sigma }:\mathbb{E}\to \mathbb{E}$ is completely continuous. By the homotopy invariance of topological degree, it follows that where I is the unit operator. By the nonzero property of the Leray-Schauder degree type, the equation $u=Tu$ has at least one solution in ${\mathcal{B}}_r$. That is, the boundary value problem (1.1)-(1.2) has at least one solution in $[0,1]$. □

Consider the following boundary value problem for fractional Langevin equations: Here we take $\alpha =1/2,\gamma =3/2,\lambda =1/4,{\beta }_i=1/2^i,{\xi }_i=1/3^i$ and $\mathbf{A}=1.64833$ and $\mathbf{B}=0.400111$ and $\eta <(1-\mathbf{B})/\mathbf{A}=0.363938$.